Just to avoid possible confusion, let me correct a typo (at step [2] in the example below). Apologies!
-----FW: <xfmail.111023084327.ted.hard...@wlandres.net>----- Date: Sun, 23 Oct 2011 08:43:27 +0100 (BST) Sender: r-help-boun...@r-project.org From: (Ted Harding) <ted.hard...@wlandres.net> To: r-help@r-project.org Subject: Re: [R] symmetric matrix multiplication On 23-Oct-11 07:00:07, Daniel Nordlund wrote: >> -----Original Message----- >> From: r-help-boun...@r-project.org >> [mailto:r-help-boun...@r-project.org] >> On Behalf Of statfan >> Sent: Saturday, October 22, 2011 10:45 PM >> To: r-help@r-project.org >> Subject: [R] symmetric matrix multiplication >> >> I have a symmetric matrix B (17x17), and a (17x17) square matrix A. >> If do >> the following matrix multiplication I SHOULD get a symmetric matrix, >> however >> i don't. The computation required is: >> >> C = t(A)%*%B%*%A >> >> here are some checks for symmetry >> > (max(abs(B - t(B)))) >> [1] 0 >> > C = t(A)%*%B%*%A >> > (max(abs(C - t(C)))) >> [1] 3.552714e-15 >> >> Any help on the matter would be very much appreciated. > > Welcome to the world of floating-point calculation on > finite precision computers. You need to read R FAQ 7.31. > Your maximum difference is for all intents and purposes > equal to zero. > > Hope this is helpful, > Dan > > Daniel Nordlund > Bothell, WA USA In addition to Dan's comment, let me point out that you can convert your very nearly symmetric matrix C to an exactly (even by R's finite-precision standards) symmetric matrix by using (C + t(C))/2. The result will differ from the original matrix C by similar "for all intents and purposes zero" amounts. Here is an example, using 4x4 matrices: ##[1]: The symmetric matrix B: B <- matrix( c( 1.1, 1.2, 1.3, 1.4, 1.2, 2.2, 2.3, 2.4, 1.3, 2.3, 3.3, 3.4, 1.4, 2.4, 3.4, 4.4), byrow=TRUE, nrow=4 ) ##[2]: The non-symmetric matrix B: [OOPS! Typo!!] ##[2]: The non-symmetric matrix A: A <- matrix( c( 1.1, 1.2, 1.3, 1.4, 2.1, 2.2, 2.3, 2.4, 3.1, 3.2, 3.3, 3.4, 4.1, 4.2, 4.3, 4.4), byrow=TRUE, nrow=4 ) ##[3]: An allegedly symmetric matrix C1 (constructed ## like your C): C1 <- t(A)%*%B%*%A ##[4]: But it isn't exactly symmetric: max(abs(C1 - t(C1))) # [1] 5.684342e-14 ##[5]: So construct an exactly symmetric version: C2 <- (C1 + t(C1))/2 ##[6]: Check that it is exactly symmetric: max(abs(C2 - t(C2))) # [1] 0 ##[7]: And check how close it is to the original C1: max(abs(C2 - C1)) # 1] 5.684342e-14 Hoping this helps! Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@wlandres.net> Fax-to-email: +44 (0)870 094 0861 Date: 23-Oct-11 Time: 08:43:18 ------------------------------ XFMail ------------------------------ ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. --------------End of forwarded message------------------------- -------------------------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@wlandres.net> Fax-to-email: +44 (0)870 094 0861 Date: 23-Oct-11 Time: 08:52:38 ------------------------------ XFMail ------------------------------ ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
